If the first three terms in the expansion of ${(1 + a x)}^{n}$ in ascending powers of $x$ are $1 + 12 x + 64 x^{2}$ , find $n$ and $a$ .

Open in App

Solution

General term is $n C_{k} a^{n - k} b^{k}$
$a = 1, b = a x$
first term is k=0 is $1$
second term k=1 is $n . a x = 12 x => n a = 12$
third term k=2 is $n C_{2} a^{2} x^{2} = 64 x^{2} => \frac{n (n - 1)}{2} . a^{2} = 64$
$=> n^{2} a^{2} - a^{2} n = 128$ but $a n = 12$ so $a^{2} n = 16$
now $\frac{a^{2} n}{a n} = \frac{16}{12} => a = \frac{4}{3}$
therefore $n = 9$

Suggest Corrections

Similar questions

In the expansion of $(1 + a x)^{n}$ , the first three terms are $1 + 12 x + 64 x^{2}$ , then n=

(1 + a x)^{n}

1, 8 x,

24 x^{2}

a =

3

(1 + x + x^{2})^{10}

x

Find a,b and n in the expansion of $(x + b)^{n}$ if the first three terms in the expansion are 729, 7290 and 30375 respectively.

8^{th}

{(3 x + \frac{2}{3 x^{2}})}^{12}

\frac{a}{x^{n}},

(a - n)

Join BYJU'S Learning Program